Let $f(x, y) = \cos(y) - \sin(x)$ and $g(t) = (-2t, 4t)$. $h(t) = f(g(t))$ $h'\left( \dfrac{\pi}{3} \right) = $
Solution: Formula The multivariable chain rule says that $\dfrac{dh}{dt} = \nabla f(g(t)) \cdot g'(t)$. The $g'(t)$ part is how much a change in $t$ will cause the input to $f$ to move, and the $\nabla f(g(t))$ part is how much $f$ will change in response to this update to its input. [What's the intuition behind the formula?] Applying the formula We want to find $h'\left( \dfrac{\pi}{3} \right) = \nabla f(g(4)) \cdot g'(4)$. $\begin{aligned} &g\left( \dfrac{\pi}{3} \right) = \left( \dfrac{-2\pi}{3}, \dfrac{4\pi}{3} \right) \\ \\ &g'\left( \dfrac{\pi}{3} \right) = (-2, 4) \\ \\ &\nabla f = (-\cos(x), -\sin(y)) \\ \\ &\nabla f\left(g \left( \dfrac{\pi}{3} \right) \right) = \nabla f \left( \dfrac{-2\pi}{3}, \dfrac{4\pi}{3} \right) = \left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right) \end{aligned}$ Substituting: $\begin{aligned} h'\left( \dfrac{\pi}{3} \right) &= \left( \dfrac{1}{2}, \dfrac{\sqrt{3}}{2} \right) \cdot (-2, 4) \\ \\ &= -1 + 2\sqrt{3} \end{aligned}$ Answer $h'\left( \dfrac{\pi}{3} \right) = -1 + 2\sqrt{3}$